Q44 / 160 · WBJEE 2012 · Mathematics

The value of $\sum_{r=2}^{\infty} \frac{1+2+\quad+(r-1)}{r !}$

  • A
    $e$
  • B
    $2 e$
  • Correct answer
    C
    $\frac{e}{2}$
  • D
    $\frac{3 e}{2}$

Solution

$\sum_{r=2}^{\infty} \frac{(r-1) r}{2 r !}=\sum_{r=2}^{\infty} \frac{1}{2(r-2) !}$ $=\frac{1}{2}\left[\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\quad \infty\right]=\frac{1}{2} e$

Hence, the correct answer is option: 'C'.

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